What will be the result if 100 mlL of 0.06 M Mg(NO_(2))_(2) is added to 50 ml. of 0.06 MNa_(2) C_(2) O_(4) ?[ Ksp of MgC_(2) O_(4) = 8.6 xx 10^(-5)]

What will be the result if 100 mlL of 0.06 M Mg(NO_(2))_(2) is added t

1.	What will be the result if 100 mlL of 0.06 M Mg(NO_(2))_(2) is added to 50 ml. of 0.06 MNa_(2) C_(2) O_(4) ?[ Ksp of MgC_(2) O_(4) = 8.6 xx 10^(-5)]
Answer» A precipitate will not be formed A precipitate will form and an excess of `Mg^(2+) ` ions will remains in the solution A precipitate willform and an excess of `C_2O_4^(2-) ` ions will remain in the solution A precipitate will form but neither ion is present in excess Solution :`[Mg^(+2) ] =0.06 M rArr [C_2O_4^(-2) ] =0. 06 M` But after mixing ` M_1V_1 = M_2V_2 rArr0. 0 6 xx 100 = M_2 xx 150` ` [Mg^(+2)] =0.04 M rArr M_1 V_1 =M_2 V_2` ` 0.06 xx 50 = M_2 xx 150 rArr [C_2 O_4^(-2) ]= 0.02 M` ` {:(Mg^(+2)+, C_2O_4^(-2)hArr, MgC_2O_4),( 0.04 M , 0.02M , 0) ,( 0.02 , -,0.02M), (0.02M , , ):}` ` I.P =[Mg^(+2) ] [C_2O _4^(-2)]` ` =(4XX 10 ^(-2) ) (2xx 10 ^(-2)) = 8 xx 10 ^(_4) ` ` thereforeK_(sp)LT I. P` , ppt will from